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Question

What is the solution of the linear-quadratic system of equations?
{ y = x^2 + 5x -3
{ y - x = 2

Answer 1

To solve the linear-quadratic system of equations, we need to find the values of x and y that satisfy both equations.

First, let's solve the second equation for y:
y - x = 2
y = x + 2

Now we substitute this expression for y into the first equation:
x^2 + 5x - 3 = x + 2

Simplifying the equation:
x^2 + 4x - 5 = 0

To solve this quadratic equation, we can either factor it or use the quadratic formula. Let's use the quadratic formula:

x = (-b ± √(b^2 - 4ac)) / (2a)
a = 1, b = 4, c = -5

x = (-4 ± √(4^2 - 4(1)(-5))) / (2(1))
x = (-4 ± √(16 + 20)) / 2
x = (-4 ± √(36)) / 2
x = (-4 ± 6) / 2

This gives us two possible values for x:
x = (-4 + 6) / 2 -> x = 1
x = (-4 - 6) / 2 -> x = -5

Now substitute these values back into the second equation to find the corresponding y-values:
For x = 1:
y = 1 + 2 -> y = 3
So one solution is (1, 3).

For x = -5:
y = -5 + 2 -> y = -3
So another solution is (-5, -3).

Therefore, the solution of the linear-quadratic system of equations is: (1, 3) and (-5, -3).